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Calculus1 22 Online
OpenStudy (anonymous):

Taylor series. I'm looking for the first few coefficients (see below). I got the first two right, then I broke it. :P

OpenStudy (anonymous):

\[\frac{3x}{1+x} = \sum_{n=0}^{\infty} c_nx^n\]

OpenStudy (anonymous):

I did this first :\[c_1=\frac{f'(x)}{1!}=3\] \[c_2=\frac{f''(x)}{2!}=-3\]So far so good. Now: \[c_3=\frac{f'''(x)}{3!}\neq\frac{3}{2}\] Am I just making an arithmetic mistake, or am I misapplying the formula?

hartnn (hartnn):

i think you did not calculate f'''(x) correctly, what did u get f'''(x) as ?

hartnn (hartnn):

because i am getting it as +3

OpenStudy (anonymous):

I had\[\frac{f'''3}{3!} = \frac{9(1+x)^{-4}}{3!}\]

hartnn (hartnn):

hmm, actually, \( f'''(x) = 18 (1+x)^{-4}\)

OpenStudy (anonymous):

Oopsie. :)

hartnn (hartnn):

did you find your error? or still need help

OpenStudy (anonymous):

Looking.

OpenStudy (anonymous):

Ok, yeah, that was it. Thanks!

hartnn (hartnn):

welcome ^_^

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